Derivative Error and Lagrange

In summary, the function f(t) = ln (1 +2x) has an nth derivative, f^(n) (x), given by (-1)^n+1 * 2^n /n * n! / (1 + 2x)^n. The maximum value of |f^(n+1)| on the interval -0.25 <= x <= 0.25 can be found using the Lagrange error bound equation for Taylor polynomials. To find the maximum value, you need to minimize the denominator in absolute value, and an upper bound can be used for this calculation. The error, |f(x) - P_n(x)|, can be expressed as M / (n+1)! * |x
  • #1
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Consider the function f(t) = ln (1 +2x)

Give a formula for f^(n) (x) [**the nth derivative] valid for all n >= 1 and find an upper bound for | f^(n) (x) | on the interval -0.25 <= x <= 0.25.
[ the error ].

I found the nth derivative to be

f^(n) (x) = (-1)^n+1 * 2^n /n * n!
--------------------------------
(1 + 2x)^n

so for
first derivative = 2 / 1+2x
second " " = -4 / (1+2x)^2
third " " = 16 / (1+2x)^3
etc.

now for the error i kno there is a lagrange error bound equation for taylor polynomials, but the question isn't for the taylor polynomial, only the "derivative generator"

i kno the max |f^(n+1)| <= M on an interval

so i just need help dealing with only the derivative error and i also want to kno how to find the M value in general with Taylor polynomials (not part of the above question)

where |f(x) - P_n(x)| <= M / (n+1)! * |x-a|^(n+1)
for interval between a and x
 
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  • #2
You are essentially done when you minimize the denominator in absolute value. Note, only "an upper bound" is asked for: You have room to over estimate.
 

1. What is a derivative error?

A derivative error, also known as approximation error, is the difference between the exact value of a derivative and its approximate value obtained through numerical methods or calculations. It is caused by the limitations of the method used and can be reduced by using more accurate methods or increasing the number of calculations.

2. How is derivative error calculated?

Derivative error is calculated by taking the absolute value of the difference between the exact derivative and the approximate derivative, divided by the exact derivative. This value is usually expressed as a percentage and is known as the relative error. The lower the relative error, the more accurate the approximation.

3. What is Lagrange interpolation?

Lagrange interpolation is a mathematical method used to approximate a function by fitting a polynomial curve through a set of known data points. It works by constructing a unique polynomial of degree n-1, where n is the number of data points, that passes through each point. This polynomial can then be used to estimate the value of the function at any point within the given interval.

4. What is the Lagrange error bound?

The Lagrange error bound is a theoretical upper bound for the error in using Lagrange interpolation to approximate a function. It is given by the formula: E = M(x-x0)(x-x1)...(x-xn)/n! where M is the maximum value of the n+1 derivative of the function within the given interval. This bound can be used to estimate the accuracy of the approximation and determine the number of data points needed for a desired level of accuracy.

5. How can derivative error and Lagrange interpolation be applied in real-life situations?

Derivative error and Lagrange interpolation are commonly used in various fields such as engineering, physics, and finance. They can be used to approximate complex functions and obtain important parameters or values that are difficult to calculate analytically. For example, they can be used to model and predict stock prices, design efficient and stable structures, and simulate physical systems. They are also used in numerical methods for solving differential equations, which are essential in many real-world applications.

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